Answered

Westonci.ca makes finding answers easy, with a community of experts ready to provide you with the information you seek. Get quick and reliable solutions to your questions from a community of experienced professionals on our platform. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.

A woman of mass 55 kg stands on the rim of a frictionless merry-go-round of radius 2.0m and rotational inertia 1250 kg-m2 that is not moving. She throws a rock of mass 350g horizontally in a direction that is tangent to the outer edge of the merry-go-round. The speed of the rock, relative to the ground, is 2.0m/s. Afterward, what are (a) the angular speed of the merry-go-round and (b) the linear speed of the woman

Sagot :

okpalawalter8

Answer:

a)   [tex]\omega=9.10*10^{-4}rad/sec[/tex]

b)   [tex]V=1.8217*10^{-3}[/tex]

Explanation:

From the question we are told that:

Mass of woman [tex]M_w=55kg[/tex]

Radius of merry go round [tex]r=2.0m[/tex]

Rotational inertia [tex]i= 1250 kg-m2[/tex]

Mass of rock [tex]M_r=350g \approx 0.350kg[/tex]

Speed of rock [tex]V_r=2.0m/s[/tex]

Tangent angle to the outer edge [tex]\theta=1[/tex]

a)

Generally the equation for conservation of momentum is mathematically given by

[tex]M_r(ucos\theta)r=(I+M_wr^2)\omega[/tex]

[tex]0.350(2.0cos1)(2.0)=(1250+(55)(2.0)^2)\omega[/tex]

[tex]1.3998=1470\omega\\\omega=\frac{1.339}{1470}[/tex]

[tex]\omega=9.10*10^{-4}rad/sec[/tex]

b)

Generally the equation for linear speed V is mathematically given by

[tex]V=r\omega\\V=2.0*9.10*10^{-4}[/tex]

[tex]V=1.8217*10^{-3}[/tex]

We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.